An elevator starts from rest with a constant
upward acceleration and moves 1 m in the first
1.7 s. A passenger in the elevator is holding a
6.6 kg bundle at the end of a vertical cord.
What is the tension in the cord as the elevator accelerates? The acceleration of gravity
is 9.8 m/s^2
To find the tension in the cord, we need to consider the forces acting on the bundle. There are two forces acting on the bundle: the tension force in the cord and the force of gravity.
Let's analyze the situation. The elevator is accelerating upwards, so its acceleration is positive. The acceleration of gravity is downward, so it's negative.
Since the passenger is holding the bundle, the vertical acceleration experienced by the bundle will be the same as the elevator's acceleration.
Now, we can use Newton's second law to relate the forces and the acceleration:
ΣF = ma
Where:
ΣF is the net force acting on the bundle,
m is the mass of the bundle, and
a is the acceleration of the elevator.
The forces acting on the bundle are the tension force (T) and the force of gravity (mg), where g is the acceleration due to gravity (9.8 m/s^2). The force of gravity acts downward, so it will be negative.
Therefore, the net force on the bundle is:
ΣF = T + (-mg) = ma
Now, let's substitute the given values into the equation:
T - mg = ma
T = ma + mg
Since the acceleration of the elevator is given, we can substitute it into the equation:
T = (6.6 kg)(a) + (6.6 kg)(9.8 m/s^2)
Finally, calculate the tension in the cord using the given values.